Question

I think there is a mistake in the solution of exercise 4G-5 in this problem set : http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/unit-4-techniques-of-integration/part-c-parametric-equations-and-polar-coordinates/problem-set-11/MIT18_01SC_pset4prb.pdf

Here is the given solution file : http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/unit-4-techniques-of-integration/part-c-parametric-equations-and-polar-coordinates/problem-set-11/MIT18_01SC_pset4sol.pdf

The curve is y = x^2 with 0 <= x <= 4 ("<=" means "inferior or equal to"). The solution integrate with respect to y. In this case, the limits should be 0 and 16, not 0 and 2.

Is this correct?

Answer 1

Yes, you are correct. The limits 0 to 2 are wrong.
If we integrated over x, we would do this:
y= x^2
dy/dx = 2x
(dy/dx)^2 = 4x^2
and
\[ ds = \sqrt{1+4x^2} \ dx\]
and the surface area formed by revolving the arc is
\[ 2 \pi \int_0^4 \sqrt{1+4x^2}\ x \ dx \]
or integrate over y using 0 to 16
either way, we get
\[ \frac{\pi}{6}\left( 65^\frac{3}{2} -1\right) \approx 273.87 \]

Answer 2

Yes! I actually first tried integrating over x as you just did, found the result you wrote, and that lead me to search a mistake in the answer. Thank you once again phi. :)